If 55034.175 kJ of heat are transferred to 150 kg of ice at a temperature of -12.15 degrees C, calculate the temperature of the resulting water.

I know my formulas for sensible and latent heat but don't know how to approach this.

Q=55034175 J

c(ice)=2060 J/kg•℃
λ=330000 J/kg
c(water)=4183 J/kg•℃

Q=Q₁+Q₂+Q₃
Q=c(ice)m(12.15-0)+λm+c(water)m(t-0),
Q/m=12.15•c(ice) +λ+c(water)•t.
Solve for “t”

To solve this problem, you will need to consider both the sensible heat and the latent heat involved in the phase change from ice to water.

First, let's calculate the sensible heat required to raise the temperature of the ice from -12.15°C to 0°C.

The formula for sensible heat is:
Q = m * c * ΔT

Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

In this case:
m = 150 kg (mass of the ice)
c = 2.09 kJ/kg°C (specific heat capacity of ice)
ΔT = 0°C - (-12.15°C) = 12.15°C

Plugging the values into the formula:
Q = 150 kg * 2.09 kJ/kg°C * 12.15°C = 3724.92 kJ

So, 3724.92 kJ of heat is required to raise the temperature of the ice to 0°C.

Next, we need to determine the amount of heat needed to convert the ice at 0°C to water at 0°C. This is known as latent heat.

The latent heat formula is:
Q = m * L

Where:
Q is the heat transferred
m is the mass of the substance
L is the latent heat of fusion

In this case:
m = 150 kg (mass of the ice)
L = 333.55 kJ/kg (latent heat of fusion of ice)

Plugging the values into the formula:
Q = 150 kg * 333.55 kJ/kg = 50032.5 kJ

So, 50032.5 kJ of heat is required to convert the ice at 0°C to water at 0°C.

Now, we can add up the sensible heat and latent heat to get the total heat transferred:
Total heat transferred = Sensible heat + Latent heat
Total heat transferred = 3724.92 kJ + 50032.5 kJ = 53757.42 kJ

Finally, we need to calculate the temperature of the resulting water.

Q = m * c * ΔT

In this case:
m = 150 kg (mass of the water)
c = 4.18 kJ/kg°C (specific heat capacity of water)
ΔT = ?

Rearranging the formula:
ΔT = Q / (m * c)
ΔT = 53757.42 kJ / (150 kg * 4.18 kJ/kg°C) = 81.770°C

Therefore, the temperature of the resulting water is 81.770°C.